If cot left(frac{A}{2}right)=sqrt{frac{1+a}{1 | vedclass

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(B) Given: $\cot \left(\frac{A}{2}ight)=\sqrt{\frac{1+a}{1-a}} \cot \left(\frac{	heta}{2}ight)$ Squaring both sides: $\cot^2 \left(\frac{A}{2}i

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$\frac{\cos A-a}{1-a \cos A}$

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(B) Given: $\cot \left(\frac{A}{2}ight)=\sqrt{\frac{1+a}{1-a}} \cot \left(\frac{	heta}{2}ight)$ Squaring both sides: $\cot^2 \left(\frac{A}{2}ight) = \left(\frac{1+a}{1-a}ight) \cot^2 \left(\frac{	heta}{2}ight)$ Using $\cot^2 \left(\frac{x}{2}ight) = \frac{1+\cos x}{1-\cos x}$,we get: $\frac{1+\cos A}{1-\cos A} = \left(\frac{1+a}{1-a}ight) \frac{1+\cos 	heta}{1-\cos 	heta}$ $\frac{1+\cos 	heta}{1-\cos 	heta} = \frac{(1+\cos A)(1-a)}{(1-\cos A)(1+a)} = \frac{1-a+\cos A-a \cos A}{1+a-\cos A-a \cos A}$ Applying Componendo and Dividendo: $\frac{(1+\cos 	heta)+(1-\cos 	heta)}{(1+\cos 	heta)-(1-\cos 	heta)} = \frac{(1-a+\cos A-a \cos A)+(1+a-\cos A-a \cos A)}{(1-a+\cos A-a \cos A)-(1+a-\cos A-a \cos A)}$ $\frac{2}{2 \cos 	heta} = \frac{2-2a \cos A}{2 \cos A-2a}$ $\frac{1}{\cos 	heta} = \frac{1-a \cos A}{\cos A-a}$ Therefore,$\cos 	heta = \frac{\cos A-a}{1-a \cos A}$

If $\cot \left(\frac{A}{2}\right)=\sqrt{\frac{1+a}{1-a}} \cdot \cot \left(\frac{\theta}{2}\right)$,then $\cos \theta=$

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